c program prints same address for strings, unexpected output of c program

Question

I was trying following code.

           #include <stdio.h>


           int main() {
                   char *s2 = "Test";
                   char *s3 = "Test";


                   printf("\n%u",s2);
                   printf("\n%u",s3);

                   return 0;
           }

When I executed this code I got following output.

           4196492
           4196492

Why the output is so.? It prints same address for s2 and s3

You have five `printf`s and only two lines of output. What is your output, really? — Robᵩ, Oct 19 '13 at 04:48
Related Question: See http://stackoverflow.com/questions/19333602/get-address-of-a-string-constant-in-c/19334244#19334244 — smRaj, Oct 19 '13 at 04:57
"It prints same address for s2 and s3" -- Yes, so? Why would you expect them to be different? — Jim Balter, Oct 19 '13 at 05:26

score 4 · Accepted Answer · edited Jun 29 '16 at 19:54

In most compilers, when a string value is directly assigned to a pointer, it’s stored in a read-only block (generally in data segment).

This is shared among functions.

when you say char *str = "Test";

"Test" is stored in a shared read-only location, but the pointer str is stored in a read-write memory and str is pointing to that memory location in the read-only location. When you assign the same string to a different pointer, say char* str2 = "Test";. The str2 pointer will point to the same address from the read only memory location, which is why the output is same.

The following code will fail; the program will crash, because you are trying to change the read only string:

int main() {
    char *str; 
    str = "Test";     // Stored in read only part of data segment
    *(str+1) = 'n'; // Trying to modify read only memory. segfault
    getchar();
    return 0;
}

Creating a string using a char array is different from creating a string with a char pointer; the char array will be created just like other types of arrays in C.

For example, If str[] is an auto variable then string is stored in the stack segment. If str[] is a global or static variable then stored in the data segment.

int main() {
    char str[] = "Test";  // Stored in stack segment like other auto variables
    *(str+1) = 'n';   //This is fine
    getchar();
    return 0;
}

The following code clarifies things:

#include <stdio.h>
char s1[] = "Test"; //Global in data segment

int main() {
    char *s2 = "Test"; //Read only 
    char *s3 = "Test"; //pointed to a string in a Read only memory 
    char s4[] = "Test"; //stack segment
    char s5[] = "Test"; //another array in stack segment

    printf("\n%p",s1); 
    printf("\n%p",s2);
    printf("\n%p",s3);
    printf("\n%p",s4);
    printf("\n%p",s5);

    return 0;
}

this will print something like this.

So you can see that addresses of s1, s4 and s5 are different.

score 0 · Answer 2 · edited Jun 29 '16 at 19:54

0

Because a String is automatically stored as a reference type in memory, It's treated as an address itself. If you store "TEST" as another pointer, that pointer will point to the same address where "TEST" was stored previously.

edited Jun 29 '16 at 19:54

ragingasiancoder

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answered Oct 19 '13 at 04:58

SANDEEP

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c program prints same address for strings, unexpected output of c program

2 Answers2